proof writing - If $n$ is a positive integer and is not a perfect square

If $n$ is a positive integer and is not a perfect square, how do you prove that $n^{1/2}$ is irrational?

Answer

Since $n$ is not a perfect number, there exists at least one prime number $p$ such that
$$n=p^\alpha q$$
where $q\in\mathbb N$ is coprime to $p$ and $\alpha\ge 1$ is odd.

Now, suppose that $n$ is a rational number, namely,
$$n^{1/2}=\frac{b}{a}$$

where $a,b$ are natural numbers and they are coprime to each other.

Then, we have
$$n=\frac{b^2}{a^2}\Rightarrow a^2n=b^2.$$

By the fundamental theorem of arithmetic, this implies that the number of $p$ in the left hand side is odd, and that the number of $p$ in the right hand side is even. This is a contradiction.

Hence, $n^{1/2}$ is an irrational number.